Question

A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.

Example 1:

Input: S = "ababcbacadefegdehijhklij" Output: [9,7,8] Explanation: The partition is "ababcbaca", "defegde", "hijhklij". This is a partition so that each letter appears in at most one part. A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.

Note:

1. S will have length in range [1, 500].
2. S will consist of lowercase letters ('a' to 'z') only.

Difficulty:Medium

Category:Two-Points, Greedy

Analyze

分析参考博客: 花花酱 LeetCode 763. Partition Labels

Solution

Solution 1: Brute Force

Time complexity: O(n^2), Space complexity: O(1)

class Solution {
 public:
  vector<int> partitionLabels(string S) {
    vector<int> ans;
    size_t start = 0, end = 0;
    for (size_t i = 0; i < S.length(); ++i) {
      // Find the las element for the first element.
      end = max(end, S.find_last_of(S[i]));
      if (end == i) {
        ans.emplace_back(end - start + 1);
        start = end + 1;
      }
    }
    return ans;
  }
};

Solution 2: Greedy

Time complexity: O(n), Space complexity: O(26/128)

class Solution {
 public:
  vector<int> partitionLabels(string S) {
    vector<int> last_index(128, 0);
    for (int i = 0; i < S.length(); ++i) last_index[S[i]] = i;
    vector<int> ans;
    int start = 0, end = 0;
    for (int i = 0; i < S.length(); ++i) {
      // Find the las element for the first element.
      end = max(end, last_index[S[i]]);
      if (end == i) {
        ans.emplace_back(end - start + 1);
        start = end + 1;
      }
    }
    return ans;
  }
};